Calculating mAH of a group of batteries connected in series

[CNYCacher]

Is it the sum of all the batteries in the pack?

[drewtam]

I think your right. Since the batteries are in series the voltage is also additive, so the actual battery life will be short but the amps put out in that hour will be high. Theoretically speaking.

Are you also accounting for the internal resistance? Because that increases as a function of the current.

Drew

[CNYCacher]

The circuit they are connected to is a low-draw circuit powered by a 5-volt regulator.  It is drawing 5mA at present.  The batteries are 1000mAH.  I am wondering if I should expect it to run for approximately 1000/5 hours or (8*1000)/5 hours.

[Hugh Damright]

I think it is still 1000 mah. If the batteries were in parallel then you would increase the mah and not the voltage ... if in series then you would increase the voltage and not the mah.

[Firethorn]

I think it is still 1000 mah. If the batteries were in parallel then you would increase the mah and not the voltage ... if in series then you would increase the voltage and not the mah.

Bingo.  In series the amperage stays the same, voltage is the sum, as is wattage(the product of the two).

1 AA NiMH, 1000 mAh, 1.2V(nominal)

In series:
2 AA NiMH, 2.4V, 1000 mAh

In parallel:
2 AA NiMH, 1.2V, 2000 mAh

Any particular reason you have 3 more batteries than you need?  If you're running it into a 5volt regulator, you should want 5 batteries, giving you 6 volts.  4 might be enough(4.8V).

If you want longer runtime you'd want to hook up more batteries in parallel, or switch out to larger rechargeables.

[CNYCacher]

Two reasons for using 8 cells.
8-cell holders are easy to find
As the battery dies, it's voltage will begin to drop, but the circuit will run as normal until the output of the 8 cells together (normally 12V), goes down to below 5V.  It is my aim to keep it running as long as possible.

[Thor]

Just remember, you're probably dumping seven volts or so straight to ground with the regulator. Of course that all hinges on the type of regulator and the circuit diagram.

[CNYCacher]

It's not a zener diode-based regulator

[Firethorn]

Two reasons for using 8 cells.
8-cell holders are easy to find
As the battery dies, it's voltage will begin to drop, but the circuit will run as normal until the output of the 8 cells together (normally 12V), goes down to below 5V.  It is my aim to keep it running as long as possible.

Are you using Alkalines or rechargables?  with the talk of 1k mAh I assumed you were talking about AA/AAA rechargeables, which have a nominal rating of 1.2 volts per cell, whereas standard non-rechargeable cells are 1.5.

So yes, it'd normally be 12 volts if you're using disposables, but only 9.6 if you're using rechargeables.

And NiCD and NiMH cells don't drop much in voltage until they're almost dead anyways.

If you really want maximum runtime, you have to remember what Thor said - most regulators will simply dump the excess voltage as heat, resulting in a circuit taking around double the juice it'd actually need. 

That's why I'd recommend dumping the AA batteries and going with C/D cells if you want extreme runtime.  Heck, a 4 pack of alkaline D cells would give you plenty of runtime.  Each one of them, for a circuit with this low of a draw, would be around 9k mAh, or around 2.5 months.  Some research says 15k, which would be around 'once a season'.  At that low of a draw, you seriously have to worry about the self discharge rate of the battery, so rechargables are not the best solution.

[CNYCacher]

Alkalines, and the 5mA was measured between the battery pack and the voltage regulator.

So basically, this thing is gonna shut off within 10 days. . .
crap

[Thor]

I'd try what Firethorn suggested and just drop the voltage regulator. Without schematics, it's really tough to figure out the total current draw, but the regulator is going to add to it and the excess voltage HAS to go somewhere.

[Firethorn]

I'd try what Firethorn suggested and just drop the voltage regulator. Without schematics, it's really tough to figure out the total current draw, but the regulator is going to add to it and the excess voltage HAS to go somewhere.

I never suggested dropping the regulator; just that adding more batteries in series; increasing voltage way beyond the 5V won't extend life.  Not knowing the details of the circuit, I don't know how sensitive it is to voltage variations.  Most devices meant to be battery powered though, are quite flexible in the voltage they'll accept. You'll get devices that take 2 batteries that'll actually work on anything from 2V - 3.5V, just as an example.

Given that you're using Alkies, 4 batteries will give you 6 volts.  So you'd double your runtime if you put two sets of 4 batteries in parallel.  It'd certainly make that regulator work less.

That'd double your life to 20 days, halving your battery usage.

The true solution is to use bigger batteries if you truly need long run times.

[CAnnoneer]

If you are committed to using all available batteries, my recommendation would be to put 2 groups in parallel, each group being a string of 4 in series. Then the output voltage would be 4.8 volts if you are using 1.2V batteries, or 6 volts, if you are using 1.5V batteries. In the former case, you can dump the regulator, because generally it is not going to be more accurate than the 0.2V difference, and it will also save you the energy that will be dissipated in it otherwise.

The recommended configuration delivers enough voltage while also maximizing efficiency in the use of available energy. I did the calculation. The exact result, assuming your fed circuit behaves like a large resistor R is:

T=(RU/E) (2r/R +1)/2

Where T is run time, U is the amh reserve of a single battery, E is the emf of a single battery, r is the internal resistance of a single battery.

By comparison, if you use a single group of four batteries in series, the result is:

T=(RU/E) (4r/R +1)/4

So, when r<<R, the recommended configuration will run twice longer than the second one, which is what we would expect.

[Thor]

Now I'm curious as to what the purpose of the project is..........

[cfabe]

Someone makes little packaged switching regulators, in a TO220 size package with a 7805 compatible pinout. I forget the company that makes them but they're about 95% efficient. If you were stuck on using the 8 cells in series, this would help you get  higher run time. But in your appliation I agree with the others and just use no regulator if you can get away without it. Wire your cells so you get an appropriate supply voltage, then parallel sets of those to get more runtime.

[Headless Thompson Gunner]

Without schematics, it's really tough to figure out the total current draw, but the regulator is going to add to it and the excess voltage HAS to go somewhere.

That's not necessarily true.  Linear regulators don't increase current draw appreciably.  Switching regulators may increase or decrease the current draw, but they don't increase the power consumption.

So what kind of voltage regulator are we using, linear or switching?  The behavior is dramatically different for each, and will have a huge impact on what kind of battery life you're going to get.  I don't think we can answer your question unless we know more about the regulator.

[Firethorn]

That's not necessarily true.  Linear regulators don't increase current draw appreciably.  Switching regulators may increase or decrease the current draw, but they don't increase the power consumption.

So what kind of voltage regulator are we using, linear or switching?  The behavior is dramatically different for each, and will have a huge impact on what kind of battery life you're going to get.  I don't think we can answer your question unless we know more about the regulator.

Excepting switching costs of course...  I doubt a regulator that small is going to be a switching one(or voltage stepping), or even if it is, I think that it would be adding a significant amount of waste.

For a circuit with that small of a power draw you're likely better off feeding it 'close enough' voltage, but I can't say what's close enough without knowing more about it.  Like I said, most stuff that takes DC can actually take a fair range of it, like 1V-1.7V.  Something that takes 5V will likely* take 4.5V-6V.  Rare is the equipment that demands 5.00 Volts.  Especially somethng that's taking 5mA@5V.  That's only 25mW, well within passive cooling range for all but the most microscopic electronics.

If you have a bench power supply that you can adjust, I'd see if it works at 4-6V.  If it does, dump the regulator unless you're paranoid and use the batteries in two parallel sets.  Since they're non-rechargeable and in a set, I don't think you have to worry much about charge reversal(the main problem when dealing with batteries in parallel).

*But no guarantees

[DustinD]

http://www.thomasdistributing.com/battery-holders.htm has battery holders. Some have 9 volt style terminals, some have two wire leads coming out of them. AA, C, D sizes, 2-10 cells.

They have Lithium ion and NiMH Rechargeable Batteries pretty cheap. They also make low discharge NiMH batteries that can hold most of their charge for over a year.

Could you add a small solar panel to help run the circuit and keep the batteries topped off? I know NiMH batteries are pretty tolerant of high trickle charge currents so you should not need a charge regulator for a reasonably small sized solar panel.

[Brad Johnson]

If you need reserve capacity get the 8 cell packs, then run the packs in parallel.  That many NiMH batteries won't be cheap, though.

Brad

[Firethorn]

Could you add a small solar panel to help run the circuit and keep the batteries topped off? I know NiMH batteries are pretty tolerant of high trickle charge currents so you should not need a charge regulator for a reasonably small sized solar panel.

That's a very good idea.

Not an endorsement, but it looks like you can get a 12V 1 Watt solar panel for $29

You would need a regulator(it'll push up to 17 volts), but the circuit would still be pretty easy and cheap.  Solar provides power and trickle charges a NiMH set, the cells provide power during dark periods.

Given that your circuit only consumes .025 watts, it's massive overkill, but I'm not seeing any smaller at the moment.

Wait:  Here we go.  6V 50mA for $17.95.

Heck, this would make it simpler...  I see no reason why that circuit couldn't also be used to power your circuit(just trade up to get the needed volts).

Given a reasonable amount of light, this system would be able to operate for years or even decades without trouble.